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TUI ĐANG GẤP CHO TÔI HỎI BÀI NÀY LỚP 2 NHA\\\\
AN CÓ 180 CÁI KẸO.BÌNH CÓ 160. HỎI BÌNH CÓ MẤY CÁI KẸO
a) Ta có: \(2.4.\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
c;=(50-49)(50+49)+(48-47)(48+47)+.............+(2+1)(2-1)
=50+49+48+............+1
=(50+1)50=2550:2=1275
d;=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)
=2^32-1
e;=(3-1)(3+1)(3^2+1)...........(3^16+1)
=(3^2-1)(3^2+1)..............(3^16+1)
=(3^16-1)(3^16+1)=3^32-1
tu tinh ket qua luy thua tao khong thua hoi dau
\(=\dfrac{8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)}{2}-\dfrac{3^{16}}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)-3^{16}}{2}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)-3^{16}}{2}\)
=-1/2
\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=3^{32}-1\)
\(\Leftrightarrow A=3^{31}-\frac{1}{2}\)
3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=(22-1)(22+1)(24+1)(28+1)(216+1)
=(24-1)(24+1)(28+1)(216+1)
=(28-1)(28+1)(216+1)
=(216-1)(216+1)
=232-1
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow\left(2^{32}-1\right)\left(2^{32}+1\right)\)
\(\Leftrightarrow2^{64}-1\)
ta có: 3=2^2-1
thay vào ta được: ''(2^2-1)(2^2+1)'' sử dụng hằng đẳng thức ta được (2^4-1)(2^4+1) tương tự ... ta được đáp án là (2^32-1)
hãy chon đúng cho mình ^ - ^
$4(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^8-1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^{16}-1)(3^{16}+1)$
$=\dfrac{3^{32}-1}{2}$